Electricity and Magnetism Transformers and Alternating Current

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1 Electricity and Magnetism Transformers and Alternating Current Lana Sheridan De Anza College Mar 16, 2018

2 Last time mutual inductance LC circuits and oscillations

3 Overview LC circuits, mechanical analogy oscillations in RLC circuits alternating current

4 Figure Energy transfer in a resistanceless, nonradiating LC circuit. The capacitor has a LC Circuits: Mechanical Analogy 32.5 Oscillations in an LC Circuit 981 a +Q i = 0 max Q max C S E L % Energy in inductor Energy in Total capacitor energy k v 0 m b i I max C q 0 L S B % Energy in inductor Energy in Total capacitor energy k m S v max c Q i 0 max Q max C S E L % Energy in inductor Energy in Total capacitor energy k v 0 m d i I max q 0 C S B L % Energy in inductor Energy in Total capacitor energy k S v max m e q i q C S E L S B % Energy in inductor Energy in Total capacitor energy k A 0 A x m S v x

5 uppose the The RLCswitch Circuits e capacitor nstant as it The switch is set first to position a, and the capacitor is charged. The switch is then thrown to position b. Of course, we can add resistors into an LC circuit. a b S e L C nal energy. this discusthin a resiscreasing in R (32.28) Figure A series RLC circuit.

6 The switch is set first to position a, and the capacitor is charged. In RLC circuits, electromagnetic energy is lost as heat in the b e L C The RLC switch Circuits: Damped Oscillations e capacitor The switch is then thrown to nstant as it position b. resistor. nal energy. a S this discusthin a resiscreasing in R Figure A series RLC (32.28) circuit. du dt = i 2 R

7 RLC Circuits: Damped Oscillations q dq C dt + L dq dt d 2 q dt 2 = ( ) dq 2 R dt Giving, d 2 q dt 2 + R L dq dt + 1 LC q = 0 The equation for a damped oscillator!

8 RLC Circuits: Mechanical Analogy Figure One example of m also act and the system energy of t ing mediu and subme in air. Afte air resistan in practice transform One co the force i tion oppos force is oft the retardi damping co ton s secon

said to be overdamped. RLC Circuits: Damped Oscillations The d 2 qq-versus-t dt 2 + R dq curve L dt + represents 1 a plot of Equation 32.31.

9 said to be overdamped. RLC Circuits: Damped Oscillations The d 2 qq-versus-t dt 2 + R dq curve L dt + represents 1 a plot of Equation LC q = 0 q Q max Charge versus LC circuit. n this way (b) Oscillong the decay in n RLC circuit. Solution a 0 t q(t) = Q max e Rt/2L cos(ω d t), where ω d = 1 LC b ( R 2L ) 2

10 Mutual Inductance Applications If there is a changing current in one coil, an emf can be induced in the other coil. The current can be transferred to a whole different circuit that is no directly connected. This can be used for wireless charging and transformers. For either of those applications to work, there must be a constantly changing current.

11 Alternating Current (AC) Alternating current (AC) power supplies are the alternative to direct current (DC) power supplies.

12 he current leads the driving emf. Alternating Current (AC) ailable at WileyPLUS ergy is thus from the generated in the resistor can be Alternating current (AC) power supplies 0 0are the π alternative 2π to 3π direct current (DC) power supplies. n 2 (v d t f). (31-68) 1 esistor, In however, an alternating is the current aver-supply, the voltage and current vary, the average sinusoidally value withof time: (a) sin u, average value of sin 2 1 u is 2 sin θ areas under the curve but sin 2 θ +1 the unshaded spaces below θ π 2π 3π (a) (31-69) θ v = V max sin(ωt) r rms, value of the current i: The sinpower 2 θ delivered to a resistive Fig. load fluctuates (a) A as plot of sin u versus u. P = +1 P max sin 2 (ωt). The average value over one cycle is zero. (b) ) (31-70) A plot of sin 2 u versus u.the average value 1 θ π 2π (b) 3π θ

13 tor can be Alternating Current (AC) (31-68) v = V max sin(ωt) 1 is the The averue of P sin = u, P max sin 2 (a) power delivered to a resistive load fluctuates as (ωt). f sin 2 1 u is 2 curve but sin 2 θ aces below (31-69) π 2π 3π 0 π 2π e current Integrating i: to find the shaded area we see that the average power delivered over Fig. a cycle will(a) be: A plot of sin u versus u. The average value over one cycle is zero. (b) (31-70) A plot of sin 2 u versus u.the average value P1 avg = P max over one cycle is. 2 2 (b) 3π θ θ

14 Alternating Current (AC) We describe the amount of current and potential difference across the circuit by its root-mean-square (RMS) value. The RMS voltage supplied is V rms = V max 2 The RMS current supplied is I rms = I max 2 Average power (resistive circuit): P = I rms ( V rms ) = 1 2 P max

15 Alternating Current (AC) Example 33.1 The voltage output of an AC source is given by the expression v = 200 sin(ωt), where v is in volts. Find the rms current in the circuit when this source is connected to a 100 Ω resistor.

16 Alternating Current (AC) Example 33.1 The voltage output of an AC source is given by the expression v = 200 sin(ωt), where v is in volts. Find the rms current in the circuit when this source is connected to a 100 Ω resistor. V rms = = 141 V

17 Alternating Current (AC) Example 33.1 The voltage output of an AC source is given by the expression v = 200 sin(ωt), where v is in volts. Find the rms current in the circuit when this source is connected to a 100 Ω resistor. V rms = = 141 V I rms = V rms R = 1.41 A

18 Transformers which is alm Transformers change V rms and I rms simultaneously, while keeping rule: Transm the average power P avg = I rms V rms constant (conservation of energy). Φ B The Ideal S The transm for efficient V N p p V s R and consum N s lower (for u sentially con Primary Secondary by Faraday The ide Fig An ideal transformer (two This works bers of turn coils via mutual wound inductance. on iron core) If thein current a basicin the first coil did not constantly In use, the transformer change circuit. (AC) An this ac would generator not work. produces current in the coil at the left (the primary). The coil v at the s = right v p (the secondary) generator w N s N p is connected to the resistive load R when

19 Summary RLC circuits alternating current Collected Homework 4 due Thursday, Mar 22. Homework Serway & Jewett: NEW: Ch 32, Probs: 45, 53, 57, 59 NEW: Ch 33, onward from page Obj. Qs: 12, 13; Conc. Qs.: 8; Probs: 1, 3, 5, 49, 51, 57

20 Appendix: Damped Oscillations Solution Derivation d 2 x dt 2 + b dx m dt + k m x = 0 Suppose an exponential function is the solution to this equation: r and B are constants. Then x = B e rt B e rt (r 2 + b m r + k m ) = 0 The exponential function is not zero for any finite t, so the other factor must be zero. We must find the roots for r that make this equation true.

21 Appendix: Damped Oscillations Solution Derivation This is called the characteristic equation The roots are: r 2 + b m r + k m = 0 ( r = b ) b 2 2m ± k 2m m This means the solutions are of the form: x = e b/(2m)t ( B 1 e iωt + B 2 e iωt) where ω = k m ( ) b 2 2m

22 Appendix: Damped Oscillations Solution Derivation This means the solutions are of the form: x = e b/(2m)t ( B 1 e iωt + B 2 e iωt) where ω = k m ( ) b 2 2m Recall that cos(x) = 1 2 (eix + e ix ). (If you haven t seen this, try to prove it using the series expansions of cosine and the exponential function.) We can write the solution as x = A e (b/2m)t cos(ωt + φ) where B 1 = A e iφ and B 2 = A e iφ.

Electricity and Magnetism Transformers and Alternating Current

Electricity and Magnetism Transformers and Alternating Current Lana Sheridan De Anza College Mar 19, 2018 Last time RLC circuits and oscillations alternating current Overview alternating current transformers